The solution in Eq.REF  can still be physical if \(\Phi _{K_m, B}(x_0) = 0\)  thus canceling the zero in the denominator. Without the pseudo-vector potential \(\mathcal {A}\) , it can be achieved when \(v_{Dirac}\)  vanishes [1]}. The same argument also applies if \(\mathcal {A}\)  is present. To prove it, consider the zero mode \(\Phi _{K_m, B}(x) = (\Phi _{K_m, t, B}(x), \ \Phi _{K_m, b, B}(x))\) . It satisfies
\(0= & \Phi _{K_m, B}^T(-x) \mathcal {D}(x) \Phi _{K_m, B}(x) \nonumber \\= & \Phi _{K_m, B}^T(x) \mathcal {D}(- x) \Phi _{K_m, B}(- x).\) 
